So I've been trying to understand a proof of the fact that $H^p(M, \mathcal F)=0$ whenever $p\geq 1$ from page 42 of the book "Principles of Algebraic Geometry" by Griffiths and Harris.

The proof is carried out for the sheaf ${\mathcal a}^{r,s}$ of $C^{\infty}$ forms of type $(r,s)$ on $U$ and there is a remark after the proof of how one can prove the result for an arbitrary fine sheaf.

According to them, a fine sheaf is a sheaf for which we have maps $\eta_\alpha:\mathcal F (U_{\alpha})\rightarrow \mathcal F (U)$ for any $U=\cup U_{\alpha}$, such that the support of $(\eta_{\alpha}\sigma)$ is contained in $U_\alpha$ and $\sum {\eta_{\alpha}} (\sigma| _{U_\alpha})=\sigma$ for $\sigma\in \mathcal F (U)$.

So is $(\eta_{\alpha}\sigma)$ a map from the manifold $M$ to $\mathbb R$? I don't understand how $(\eta_{\alpha}\sigma)$ has $M$ as its domain.

Secondly how can they write $(\eta_{\alpha}\sigma)$ when $\sigma\in \mathcal F (U)$? Isn't the domain of $\eta_\alpha$ the group ${\mathcal F }(U_\alpha)$?

Thirdly how does this formulation apply to the case when $\mathcal F$ is the sheaf ${\mathcal a}^{(r,s)}$? We would need maps $\eta_{\alpha}: {\mathcal a}^{(r,s)}(U_{\alpha})\rightarrow {\mathcal a}^{(r,s)}(U)$. But indstead the proof just uses the partion of unity $C^{\infty} $ maps $\rho_{\alpha}$ subordinate to some open cover $\{U_\alpha\}$. So it seems that they haven't used the fact that ${\mathcal a}^{r,s}$ is a fine sheaf.